hagman
Posts:
1,923
Registered:
1/29/05
|
|
Re: Proof 0.999... is not equal to one.
Posted:
May 31, 2007 3:32 PM
|
|
On 31 Mai, 14:04, chaja...@mail.com wrote:
> This has been a very educational evening for me. I would still like to
> point out that no one has refuted my proof itself or pointed to
> specific error or logical over-stepping it may contain.
>
> http://www17.brinkster.com/chajadan/Math/Proofs/Proof1.doc
>
> http://www17.brinkster.com/chajadan/Math/Proofs/Proof1.odt
>
> http://www17.brinkster.com/chajadan/Math/Proofs/Proof1.txt
>
> > --Charles J. Daniels
OK, here are some notes:
Near the bottom of page 1 you write:
Let S be the set of all real numbers in the interval [0,1]
Let T be the set of all real numbers in the interval (-1,0]
If you applied an operator '+' to these two sets that sums the
elements of both sets,
you would get: S + T = 1
You have never defined this operator '+' for intervals.
The standard use (derived from the '+' for numbers) is to take the set
of all possible sums x+y where x is in S and y is in T.
However, this would result in S + T = (-1,1].
Most of the text is so very hand-wavingly informal that it is not even
wrong.
I found one portion, however, near the top of page 2, that sounds
mathematically
correct to me:
Let x = 0.999...
x = 0.999... {1} Multiply both sides by 10:
10x = 9.999... {2} Subtract line {1} from line {2}:
9x = 9 {3} Divide both sides by 9:
x = 1
Since x = 0.999... and x = 1, it must be the case that 1 = 0.999...
Later, your arguments, especially on page 4 are totally circular.
Page 5 shows that you seem to grasp that there is a necessity to
*define*
the meaning of infinite sums since things can go wrong when
rearranging series
that are not absolutely converging.
To save my own sanity, I scrolled ahead until I found this example
of downright nonsense:
However, defining it as a limit is all but directly saying that it
may
not be exactly equal to one.
Later on
A Nested Set theorem says that a sequence of nested closed sets that
become arbitrarily small eventually contain only a single real
number.
Then they show something like this:
[0, 1], [0.9, 1], [0.99, 1], ..., [0.999..., 1]
But what about the closed set [1]? How could we have felt sure
0.999...
would be in this set as well? The nested set theorem only guaranteed
you'd get to a single value.
Indeed, you'd get only a single value AND you'd get 0.999... AND you'd
get 1. This implies 0.999...=1.
Maybe, the problem is that you have never defined what "0.999..."
means.
In this thread you claimed that it should be the sum of all elements
of the set S where S contains all numbers of the form 9/10^n for n>=0.
So the problem is shifted to defining what the sum of all elements
of an infinite set of positive rational numbers is.
Two cases arise:
a) For every integer M, we can find a finite subset of S such that
the
(clearly defined) sum of these finitely many elements is bigger
than M.
In this case, there is no hope of defining the sum of S or it is
said
to diverge.
b) For some M, it is impossible to get beyond M with a finite subset.
We might be able to define a sum here. In fact, *I* can make such
a
definition in a consistent manner.
One can show that your set S is of type b).
Thus one can hope to define sum(S) and in fact whatever you claim
0.999... to be is what you define as sum in this case.
If r is a positive rational number, one can define the set T
containing
all elements of S multiplied by r, i.e. the numbers r*9/10^n.
It should definitely be true that, whenever either sum(S) or sum(T)
is defined so is the other and the equality
sum(T) = r * sum(S)
holds, shouldn't it?
If s is a rational number not contained in S, we can define the set
U = S u {s} containing all elements of S and also s.
It should definitely be true that, whenever either sum(S) or sum(U)
is defined so is the other and the equality
sum(U) = s + sum(S)
holds, shouldn't it?
With your set S defined as above, I choose r=10 to form T and I choose
s=9
to form U.
If you look at it closely, you will note that T=U holds.
It should definitely be true that the equality
sum(U) = sum(T)
holds in this case, shouldn't it?
But then
r * sum(S) = sum(T) = sum(U) = s + sum(S)
i.e.
10 * sum(S) = 9 + sum(S).
In my math, this can be transformed to
sum(S) = 1.
Your mileage my vary, but then you have not given
adequate definitions of '+', '*', '='.
hagman
|
|
